4 Solved Problems on Computer Organization - Quiz 2 | CPSC 2105, Quizzes of Computer Architecture and Organization

Material Type: Quiz; Class: Computer Organization; Subject: Computer Science; University: Columbus State University; Term: Spring 2007;

Typology: Quizzes

Pre 2010

Uploaded on 08/04/2009

koofers-user-f3s
koofers-user-f3s 🇺🇸

10 documents

1 / 3

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
CPSC 2105 Introduction to Computer Organization
Quiz 2 Wednesday, February 14, 2007
NOTE: SHOW ALL OF YOUR WORK. YOU MUST DEMONSTRATE THAT YOU
UNDERSTAND THE PROCESS OF OBTAINING AN ANSWER. WHILE THESE CAN
BE DONE ON MANY CALCULATORS, YOU MUST DO THESE BY HAND.
1. Show the truth table for a NAND gate.
Answer:
A B AB Not (AB)
0 0 0 1
0 1 0 1
1 0 0 1
1 1 1 0
2. What does the following circuit do? Describe it using a truth table.
Answer: We begin by labeling the outputs
Here is the truth table
A B I II III C
0 0 1 1 1 0
0 1 1 1 0 1
1 0 1 0 1 1
1 1 0 1 1 0
C = A B. On the next page, I use DeMorgan’s laws to derive the expression.
Page 1 of 3 pages Answer Key 2/16/2007
pf3

Partial preview of the text

Download 4 Solved Problems on Computer Organization - Quiz 2 | CPSC 2105 and more Quizzes Computer Architecture and Organization in PDF only on Docsity!

CPSC 2105 Introduction to Computer Organization

Quiz 2 Wednesday, February 14, 2007

NOTE: SHOW ALL OF YOUR WORK. YOU MUST DEMONSTRATE THAT YOU

UNDERSTAND THE PROCESS OF OBTAINING AN ANSWER. WHILE THESE CAN

BE DONE ON MANY CALCULATORS, YOU MUST DO THESE BY HAND.

  1. Show the truth table for a NAND gate. Answer: A B AB Not (AB) 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0
  2. What does the following circuit do? Describe it using a truth table. Answer: We begin by labeling the outputs Here is the truth table A B I II III C 0 0 1 1 1 0 0 1 1 1 0 1 1 0 1 0 1 1 1 1 0 1 1 0 C = A  B. On the next page, I use DeMorgan’s laws to derive the expression.

CPSC2105 Quiz 2 2/14/ Obviously, we have C = A  B, with the truth table. A B C = A  B 0 0 0 0 1 1 1 0 1 1 1 0

F(A, B) = (1, 2) = (0, 3)